Dr Nosipho Moloto Room C206, Humphrey Raikes Building

Transcription

Dr Nosipho Moloto
Room C206, Humphrey Raikes Building
[email protected]
1
The Concept of Equilibrium
o Chemical equilibrium occurs when opposing reactions
are proceeding at equal rates.
o Can be achieved only when a chemical reaction is
reversible.
o Consider the dissociation of N2O4 to form NO2
N2O4(g)
Colourless
2NO2(g)
brown
oAt equilibrium an equilibrium mixture of N2O4 and
NO2 remains, wherein the concentrations of the gases
no longer change with time.
2
3
o Using Reaction Kinetics, the equation
N2O4 (g)
2 NO 2 (g)
o The rate laws for the forward and reverse elementary
reactions can be written as follows:
Forward Reaction: N2O4 (g)
Reverse Reaction: 2 NO 2 (g)
2 NO 2 (g) Rate
N2O4 (g) Rate
f
r
k f [N2O4 ]
k r [NO 2 ]2
Rate constants
Rate f k f [N2O4 ]
Rate r k r [NO 2 ]2
k f [N2O4 ] k r [NO 2 ]2
Rearranging gives:
kf
kr
2
[NO2 ]
[N2O4 ]
constant
5
6
Summary
• At equilibrium, the concentrations of reactants
and products no longer change with time.
• For equilibrium to occur, neither reactants nor
products escape from the system.
• At equilibrium, a ratio of concentration terms
equals a constant – the equilibrium constant.
7
The Equilibrium Constant
o Consider the Haber Process:
N2(g) + 3 H2 (g)
2 NH3 (g)
8
o In 1864 GULDBERG and WAAGE postulated the now famous
Law of Mass Action. According to the law of mass action:
Any reversible chemical reaction
aA +bB
dD + eE
will have associated with it an equilibrium constant, Kc
Kc
[D]d [E] e
[A] a [B]b
Products
Reactants
Applies irrespective of whether the reaction is elementary or
complex.
9
o Thus for the Haber Process:
2
Kc
[NH3 ]
3
[N2 ][H2 ]
Note: The equilibrium constant-expression depends only
on the stoichiometry of the reaction, not on its
mechanism.
10
Evaluating Kc
The equilibrium constant is independent of the starting
concentrations of the reactants in chemical reaction.
1
2
3
4
11
Evaluating Kc
12
Equilibrium Constants in Terms of Pressure Kp
• When both reactants and products are gases in a chemical
reaction, the equilibrium constant expression can be written
in terms of partial pressures of the gases.
• Thus for the general reaction:
aA +bB
dD + eE
Kp
(PD) d (PE) e
(PA) a (PB)b
13
Equilibrium Constants in Terms of Pressure, Kp
Example:
N2O4 (g)
Kp
2 NO 2 (g)
(PNO2)2
(PN2O4)
14
Equilibrium Constants in Terms of Pressure Kp
• Generally, the value of Kc is different from Kp for a given reaction.
• However, using the ideal-gas equation we can calculate Kp given Kc
and vice versa.
• Thus
PV = nRT, P = (n/V)RT
where n/V = mol/L = M (molarity)
For a substance A,
PA = (nA/V)RT = [A]RT
Eventually after substitution,
Kp = Kc(RT)∆n
15
The Magnitude of the Equilibrium Constant
Consider the magnitude of the equilibrium constant for the
following reaction:
CO (g) + Cl2 (g)
Kc
[COCl 2 ]
[CO ][ Cl 2 ]
COCl2 (g);
4.56 10
9
For Kc to be this large, the numerator (equilibrium
concentration of COCl2) must be much greater than the
denominator.
Here equilibrium lies to the right.
16
In general,
If K>>1 (K is large): Equilibrium lies to the right; products
predominate
If K<<1 (K is small): Equilibrium lies to the left; reactants
predominate
17
To summarize:
18
The direction of the chemical equation and K
aA + bB
cD + dD
d
Kc
e
[C] [D]
a
b
[A] [B]
If we invert the chemical equation for the reversible reaction:
cD + dD
K' c
aA + bB
[A] a [B] b
[C] c [D] d
We see that K’c = 1/Kc
19
Relating Chemical Equations and the Equilibrium Constants
Consider:
N2O4 (g)
2 NO 2 (g)
Multiply it by 2: 2N2O4 (g)
4 NO 2 (g)
4
has
Kc
[NO2 ]
[N2O4 ]2
which is the square of the equilibrium constant
for N2O4 (g)
2 NO 2 (g)
20
Relating Chemical Equations and the Equilibrium Constants
• The equilibrium constant for the reverse direction is the
inverse of that for the forward direction.
• When a reaction is multiplied by a number, the
equilibrium constant is raised to that power.
• The equilibrium constant for a reaction which is the sum
of other reactions is the product of the equilibrium
constants for the individual reactions.
• Note: Kc , Keq or Kp are symbols for the equilibrium
constant.
21
Heterogeneous Equilibria
Homogeneous equilibria – equilibria in which all reactants
and products are in the same phase (e.g., all in the gas
phase; all dissolved in the same solution).
Example: N2O4 (g)
2 NO 2 (g)
Heterogeneous equilibria – equilibria in which
some or all of the reactants and products are in
different phases.
PbCl 2 (s)
CaCO 3 (s)
Pb 2 (aq) 2 Cl - (aq)
CaO (s) CO 2 (g)
22
If a pure solid or liquid is involved in a heterogeneous
equilibrium, its concentration is not included in the
equilibrium constant expression for the reaction.
CaCO 3 (s)
Kc = [CO2]
CaO (s) CO 2 (g)
and
Kp = PCO2
23
Some examples
SnO2 (s) 2CO(g)
CaCO3 (s)
2
Zn(s) Cu (aq)
Sn(s) 2CO2 (g)
CaO(s) CO2 (g)
2
Cu(s) Zn (aq)
24
In a case where a solvent is involved as a reactant or product
in an equilibrium, its concentration is also excluded in the
equilibrium-constant expression.
Thus for:
2
3
H2O(l) CO (aq)
Kc
OH (aq) HCO3 (aq)
[OH ][HCO3 ]
2
[CO 3 ]
[H2O] is excluded from the expression.
25
Calculating Equilibrium Constants
If all equilibrium concentrations are known, one simply
substitutes these values into the equation for Kc or Kp
for the reaction.
Example 1: Calculating K when all equilibrium concentrations
are known.
Given that at 852K the equilibrium concentrations of the reactants
and products are
[SO2] = 3.61 x 10-3 mol L-1 ; [O2] = 6.11 x 10-4 mol L-1
[SO3] = 1.01 x 10-2 mol L-1 for the reaction
2SO2 (g) + O2 (g)
2SO3 (g)
26
Work out the equilibrium constant
2
Kc
[SO3 ]
[SO2 ]2 [O 2 ]
2 2
Kc
[1.01 10 ]
[3.61 10 2 ]2 [6.11 10 4 ]
= 1.28 Χ 102
Kc is large
⇒ equilibrium favours___________
27
Example 2: Calculating K when all equilibrium concentrations
are known.
Note: We are given partial pressures therefore we need to
use Kp expression.
28
Example 2:
29
If we don’t know the equilibrium concentrations of all
chemical species in an equilibrium, but we know at least
one of the concentrations, we can use stoichiometry to
work out the unknown concentrations.
Example: Calculating K from initial and equilibrium
concentrations.
Consider the equilibrium: H2 (g) + I2 (g)
2 HI (g)
Suppose the system initially contains 1.000 X 10-3 M H2 and
2.000 X 10-3 M I2 at 448oC. At equilibrium the conc of HI is
found to be 1.87 X 10-3 M. Calculate Kc for the reaction at
448oC.
30
Solution:
1
H2 (g)
Initial Conc (M)
Change in conc
Equilibrium / final
Conc.
+
I2 (g)
2 HI (g)
1.000 X 10-3 M 2.000 X 10-3 M 0 M
1.87 X 10-3 M
2
Change in [HI] = 1.87 X103- M – 0 = 1.87 X 10-3 M
3
Use coefficients in the balanced equation to relate change in [HI]
to the changes in [H2] and [I2].
31
3
4
1.87 X 10-3 mol HI
L
1 mol H2 = 0.935 X 10-3 M H
2
2 mol HI
1.87 X 10-3 mol HI
L
1 mol I2 = 0.935 X 10-3 M I
2
2 mol HI
Calculate the equilibrium concentrations of H2 and I2 using
the initial concentrations and changes.
[H2] = 1.000 X 10-3 M – 0.935 X 10-3 M = 0.065 X 10-3 M
[I2] = 2.000 X 10-3 M – 0.935 X 10-3 M = 1.065 X 10-3 M
32
4
The completed table becomes:
H2 (g)
Initial Conc (M)
Change in conc
Equilibrium / final
Conc.
+
I2 (g)
2 HI (g)
1.000 X 10-3 M 2.000 X 10-3 M 0 M
- 0.935 X 10-3M - 0.935 X 10-3M + 1.87 X 10-3 M
0.065 X 10-3 M 1.065 X 10-3 M 1.87 X 10-3 M
Therefore, using the equilibrium-constant expression, we
calculate the equilibrium constant:
2
Kc
[HI]
[H2 ][I2 ]
3 2
1.87 10
0.065 10 3 1.065 10
3
51
33
Applications of Equilibrium Constants
Predicting the Direction of Reaction
o If only initial concentrations are known, we use the
REACTION QUOTIENT expression, Q. This expression is
equal to the Keq (Kp or Kc) expression when the system is
in equilibrium.
Q = K at equilibrium
In general the quotient Q is expressed as follows for
a A (g) + b B (g)
c C (g) + d D (g)
Q = (original conc C)c (original conc D)d
(original conc A)a (original conc B)b
34
Qc
[D]d [E] e
[A] a [B]b
d
or
Qc
e
PD PE
a b
PA PB
o If Q > K then the reverse reaction must occur to reach
equilibrium (i.e., products are consumed, reactants are
formed, the numerator in the equilibrium constant
expression decreases and Q decreases until it equals K).
35
o If Q < K then the forward reaction must occur to reach
equilibrium (i.e. reactants are consumed, products are
formed, the numerator in the equilibrium constant increases
and Q increases until it equals K).
We can therefore predict the direction of the reaction
36
37
Calculating Equilibrium Concentrations
 Use similar approach used to solve problems involving
equilibrium constants.
i.e. Tabulate the initial concentrations or partial pressures,
the changes calculated and the final equilibrium
concentrations or partial pressures.
Example: For the Haber Process N2(g) + 3 H2(g)
2 NH3(g),
Kp = 1.45 X10-5 at 500°C. In an equilibrium mixture of the
three gases at 500°C, the partial pressure of H2 is 0.928 atm
and that of N2 is 0.432 atm. What is the partial pressure of
NH3?
38
• Here we are given Kp and two of three partial pressures
i.e. one unknown.
N2(g) + 3 H2(g)
0.432
Kp
(PNH3 )
0.928
2
PN2 (PH2 )
3
2 NH3(g)
x
(x)2
3
(0.432)(0.928)
1.45 10
39
5
Example: Consider the reaction H2(g) + I2(g)
2 HI(g)
In which 1L flask is filled with 1.000 mol of H2 and 2.000 mol
of I2 at 448 C. The value of Kc for the reaction is 50.5 at
448 C. What are the equilibrium concentrations of H2, I2 and
HI in mol/L?
1 Note initial concentrations in the 1L flask:
[H2] = 1.000 M and H2 = [2.000 M]
40
2
Initial Conc (M)
Change in conc
H2 (g)
1.000 M
+
I2 (g)
2 HI (g)
2.000 M
0M
Equilibrium / final
Conc.
3 For each x mol of H2 that reacts, x mol of I2 are
consumed and 2x mol of HI are produced
41
3
Initial Conc (M)
Change in conc
H2 (g)
+
1.000 M
-x
I2 (g)
2 HI (g)
2.000 M
-x
0M
+ 2x
Equilibrium / final
Conc.
4
Initial Conc (M)
Change in conc
Equilibrium / final
Conc.
H2 (g)
+
I2 (g)
2 HI (g)
1.000 M
2.000 M
0M
-x
(1.000 – x) M
-x
(2.000 – x) M
+ 2x
2x M
42
5
Kc
[HI]2
[H2 ][I2 ]
(2x)2
(1.000 x)(2.000 x)
50.5
Solve for x
4x
2
50.5(x
2
3.000x 2.000)
46.5x2 - 151x 101.0 0
ax2 + bx + c
x
( 151.5)
( 151.5)2 4(46.5)(101.0)
2(46.5)
2.323or0.935
43
Substituting into the equilibrium expressions we find that x =
2.323 gives a negative concentration, which we reject
because it is not chemically meaningful. We therefore use x =
0.935 to find the equilibrium concentrations:
[H2] = 1.000 – x = 0.065 M
[I2] = 2.000 – x = 1.065 M
[HI] = 2x = 1.87 M
Check:
Kc
[HI]2
[H2 ][I2 ]
(1.87)2
(0.065)(1.065)
51
44
Le Châtelier’s Principle
 A chemical reaction will eventually reach equilibrium,
but what happens if we change conditions such as:
 Pressure/volume
 Add/remove reactant or product
 Change Temperature
Add a catalyst
45
Example:
In the Haber Process for the production of ammonia,
based on the reversible reaction:
N2 (g) + 3 H2 (g)
2 NH3
it is observed that:
• As the total pressure increases, the amount of ammonia
present at equilibrium increases.
• As the temperature decreases, the amount of ammonia
at equilibrium increases.
46
47
Le Châtelier’s Principle States:
If a system at equilibrium is disturbed by a change in
temperature, pressure, or the concentration of one of the
components, the system will shift its equilibrium position
so as to counteract the effect of the disturbance.
48
Change in Reactant or Product Concentrations
• Consider and equilibrium mixture of N2, H2, and NH3 in the
Haber Process:
N2 (g) + 3 H2 (g)
2 NH3
• If H2 is added while the system is at equilibrium, the
system must respond to counteract the addition of H2 (by
Le Châtelier’s Principle).
• The system must consume the H2 and produce more of
the products until a new equilibrium is established.
• So, most of the additional H2 will be consumed, and [N2]
will decrease, while [NH3] will increase.
49
Note new equilibrium
50
• Adding a reactant or product shifts the position of
equilibrium away from the increase.
• Removing a reactant or product shifts the equilibrium
towards the decrease.
• To optimize the amount of product at equilibrium, we
need to flood the reaction vessel with reactant and
continuously remove product (by Le Châtelier’s Principle).
• We illustrate the application of these principles with the
industrial preparation of ammonia.
51
52
Effects of Volume and Pressure Changes
• If the volume of a system at equilibrium is decreased, the
pressure increases.
• Le Châtelier’s Principle: if pressure is increased the
position of equilibrium will shift to counteract the increase.
• That is, the position of equilibrium shifts to remove gases
and decrease pressure.
• An increase in pressure favours the direction that has
fewer moles of gas.
• In a reaction with the same number of product and
reactant moles of gas, pressure has no effect.
53
Example
54
• As volume is decreased pressure increases
• An increase in pressure (by decreasing the volume)
favours the formation of N2O4.
• The instant the pressure increases, the system is not at
equilibrium and the concentration of both gases has
increased.
• The system moves to reduce the number moles of gas
(i.e. the reverse reaction is favoured).
55
Effect of Temperature Changes
• Changes in concentration and pressure do not affect the
value of the equilibrium constant.
• The equilibrium constant is temperature dependent. i.e.
it changes in value as the temperature changes.
• For an endothermic reaction, ΔH > 0 and heat can be
considered as a reactant.
Reactants + heat
products
• For an exothermic reaction, ΔH < 0 and heat can be
considered as a product.
Reactants
products + heat
56
57
• Adding heat (i.e. heating the vessel) favours the
direction of the reaction that absorbs (removes) heat:
– if ΔH > 0, adding heat favours the forward reaction,
– if ΔH < 0, adding heat favours the reverse reaction.
• Removing heat (i.e. cooling the vessel), favours the
direction of the reaction that liberates (produces) heat:
– if ΔH > 0, cooling favours the reverse reaction,
– if ΔH < 0, cooling favours the forward reaction.
58
In summary:
 Endothermic reaction: Increasing T results in an increase
in K.
 Exothermic reaction: Increasing T results in a decrease in
K.
59
The Effect of Catalysts
• A catalyst lowers the activation energy barrier for the
reaction i.e. between the reactants and products.
• Therefore, a catalyst will decrease the time taken to
reach equilibrium.
• A catalyst does not change the composition of the
equilibrium mixture.
60
The Effect of Catalysts
Example: Consider a hypothetical reaction A
B
Less energy
required
61
Additional Problems on Gas Phase Equilibria
Problem:
2 SO3 (g)
2 SO2 (g) + O2 (g)
62
Solution:
63
2 SO3 (g)
2 SO2 (g) + O2 (g)
PV = nRT
64
65
66
67
68
End of Chapter 15
-------------------------------Chemical Equilibrium
69

Dr Nosipho Moloto Room C206, Humphrey Raikes Building

Transcription

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